Difference between revisions of "1996 AIME Problems/Problem 10"
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Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>. | Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>. | ||
The smallest positive solution of this is <math>x = \boxed{159}</math> | The smallest positive solution of this is <math>x = \boxed{159}</math> | ||
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+ | == See Also == | ||
+ | |||
+ | {{AIME box|year=1996|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA NoticeKALKDL:AJ LKRLJKEHLSLKWJLTP{TWK:}} |
Revision as of 00:48, 28 May 2020
Contents
Problem
Find the smallest positive integer solution to .
Solution
.
The period of the tangent function is , and the tangent function is one-to-one over each period of its domain.
Thus, .
Since , multiplying both sides by yields .
Therefore, the smallest positive solution is .
Solution 2
which is the same as .
So , for some integer . Multiplying by gives . The smallest positive solution of this is
See Also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
{{MAA NoticeKALKDL:AJ LKRLJKEHLSLKWJLTP{TWK:}}